How do you find the volume of the region below #y= -3x+6# and enclosed by the y-axis from 0 to 2, rotated about the x-axis?

1 Answer
Jun 11, 2015

Answer:

#V_"tot"=56/9pi.#

Explanation:

First of all, we have to understand how the region of the plane is made:
The region is determined by this system:
#{(y>=0),(y<=2),(y<=-3x+6),(x>=0):}#
And the points interested are:
#P={A(0,2), B(4/3,2), C(2,0),D(0,0)}#

Geogebra
This is the formula for calculating the volume of a solid obtained from a revolution around the x-axis: #V=pi*int_a^b f^2(x) dx#
But in this case f(x) is a composite function:
#f(x)={(y=2 if 0<=x<=4/3),(y=-3x+6 if 4/3<=x<=2):}#

So we should break the solid into two rotation solids:
a cylinder from x=0 to x=4/3 and a cone from x=4/3 to x=2:
#V_"tot"=V_"cylinder"+V_"cone"#
#V_"cylinder"=piint_0^(4/3)(2)^2dx=pi[4x]_0^(4/3)=16/3pi.#

#V_"cone"=piint_(4/3)^2(-3x+6)^2dx = piint _(4/3) ^2 (9x^2-36x+36)dx#

#V_"cone" = pi[3x^3-18x^2+36x]_(4/3)^2=pi(3*2^3-18*2^2+36*2-3(4/3)^3+18(4/3)^2-36(4/3))#

#V_"cone" = pi(24-72+72-64/9+288/9-48)=pi(-216/9-64/9+288/9)=8/9pi#

#V_"tot"=16/3pi+8/9pi=56/9pi.#